Rank List（map）-优快云博客

本文链接：https://blog.youkuaiyun.com/with_wine/article/details/119816919

该篇博客介绍了一个关于学生考试排名的问题。李明总是在每次考试后询问他在班级的排名，这使老师感到疲惫。于是老师决定提供所有学生的分数，让李明自己计算排名。博客中给出了一段C++代码示例，利用地图（map）数据结构来处理这个问题。代码读取学生名单和考试成绩，然后计算并输出李明在每次考试后的排名。博客内容涉及基础的数据结构应用和简单的算法设计。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Li Ming is a good student. He always asks the teacher about his rank in his class after every exam, which makes the teacher very tired. So the teacher gives him the scores of all the student in his class and asked him to get his rank by himself. However, he has so many classmates, and he can’t know his rank easily. So he tends to you for help, can you help him?

Input

The first line of the input contains an integer N (1 <= N <= 10000), which represents the number of student in Li Ming’s class. Then come N lines. Each line contains a name, which has no more than 30 letters. These names represent all the students in Li Ming’s class and you can assume that the names are different from each other.

In (N+2)-th line, you'll get an integer M (1 <= M <= 50), which represents the number of exams. The following M parts each represent an exam. Each exam has N lines. In each line, there is a positive integer S, which is no more then 100, and a name P, which must occur in the name list described above. It means that in this exam student P gains S scores. It’s confirmed that all the names in the name list will appear in an exam.

Output

The output contains M lines. In the i-th line, you should give the rank of Li Ming after the i-th exam. The rank is decided by the total scores. If Li Ming has the same score with others, he will always in front of others in the rank list.

Sample Input
3
Li Ming
A
B
2
49 Li Ming
49 A
48 B
80 A
85 B
83 Li Ming
Sample Output
1
2
Sponsor

map的基础应用。

#include<stdio.h>
#include<string>
#include<map>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
	int i,j,k,n,m,cnt;
	char s[100];
	while(scanf("%d",&n)!=EOF)
	{
		map<string,int>mp;
		map<string,int>::iterator it,it2;
		getchar();
          for(i=0;i<n;i++)
		  {
			  gets(s);
			  mp[s]=0;
		  }
		scanf("%d",&m);
		for(i=0;i<m;i++)
		{
			cnt=1;
			for(j=0;j<n;j++)
			{
				scanf("%d",&k);
				getchar();
				gets(s);
				mp[s]+=k;
			}
			for(it=mp.begin();it!=mp.end();it++)
			{
				 if(it->second>mp["Li Ming"]) cnt++;
			}
			printf("%d\n",cnt);
		}
	}
	return 0;
}